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Formatted question description: https://leetcode.ca/all/646.html

# 646. Maximum Length of Pair Chain (Medium)

You are given n pairs of numbers. In every pair, the first number is always smaller than the second number.

Now, we define a pair (c, d) can follow another pair (a, b) if and only if b < c. Chain of pairs can be formed in this fashion.

Given a set of pairs, find the length longest chain which can be formed. You needn't use up all the given pairs. You can select pairs in any order.

Example 1:

Input: [[1,2], [2,3], [3,4]]
Output: 2
Explanation: The longest chain is [1,2] -> [3,4]


Note:

1. The number of given pairs will be in the range [1, 1000].

Related Topics:
Dynamic Programming

Similar Questions:

## Solution 1. DP

First sort the array in ascending order.

Let dp[i] be the length of maximum chain formed using a subsequence of A[0..i] where A[i] must be used.

dp[i] = max(1, max(1 + dp[j] | pair j can go after pair i ))

// OJ: https://leetcode.com/problems/maximum-length-of-pair-chain/
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
int findLongestChain(vector<vector<int>>& A) {
sort(begin(A), end(A));
int N = A.size();
vector<int> dp(N, 1);
for (int i = 1; i < N; ++i) {
for (int j = 0; j < i; ++j) {
if (A[j][1] >= A[i][0]) continue;
dp[i] = max(dp[i], 1 + dp[j]);
}
}
return *max_element(begin(dp), end(dp));
}
};


## Solution 3. Interval Scheduling Maximization (ISM)

// OJ: https://leetcode.com/problems/maximum-length-of-pair-chain/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int findLongestChain(vector<vector<int>>& A) {
sort(begin(A), end(A), [](auto &a, auto &b) { return a[1] < b[1]; });
int e = INT_MIN, ans = 0;
for (auto &v : A) {
if (e >= v[0]) continue;
e = v[1];
++ans;
}
return ans;
}
};

• class Solution {
public int findLongestChain(int[][] pairs) {
Arrays.sort(pairs, new Comparator<int[]>() {
public int compare(int[] pair1, int[] pair2) {
if (pair1[0] != pair2[0])
return pair1[0] - pair2[0];
else
return pair1[1] - pair2[1];
}
});
int length = pairs.length;
int[] dp = new int[length];
Arrays.fill(dp, 1);
for (int i = 1; i < length; i++) {
int[] curPair = pairs[i];
for (int j = 0; j < i; j++) {
int[] prevPair = pairs[j];
if (prevPair[1] < curPair[0])
dp[i] = Math.max(dp[i], dp[j] + 1);
}
}
int max = 1;
for (int num : dp)
max = Math.max(max, num);
return max;
}
}

############

class Solution {
public int findLongestChain(int[][] pairs) {
Arrays.sort(pairs, Comparator.comparingInt(a -> a[1]));
int ans = 0;
int cur = Integer.MIN_VALUE;
for (int[] p : pairs) {
if (cur < p[0]) {
cur = p[1];
++ans;
}
}
return ans;
}
}

• // OJ: https://leetcode.com/problems/maximum-length-of-pair-chain/
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
int findLongestChain(vector<vector<int>>& A) {
sort(begin(A), end(A));
int N = A.size();
vector<int> dp(N, 1);
for (int i = 1; i < N; ++i) {
for (int j = 0; j < i; ++j) {
if (A[j][1] >= A[i][0]) continue;
dp[i] = max(dp[i], 1 + dp[j]);
}
}
return *max_element(begin(dp), end(dp));
}
};

• class Solution:
def findLongestChain(self, pairs: List[List[int]]) -> int:
ans, cur = 0, -inf
for a, b in sorted(pairs, key=lambda x: x[1]):
if cur < a:
cur = b
ans += 1
return ans

############

class Solution(object):
def findLongestChain(self, pairs):
"""
:type pairs: List[List[int]]
:rtype: int
"""
tails = []
for start, end in sorted(pairs):
idx = bisect.bisect_left(tails, start)
if idx == len(tails):
tails.append(end)
else:
tails[idx] = min(tails[idx], end)
return len(tails)


• func findLongestChain(pairs [][]int) int {
sort.Slice(pairs, func(i, j int) bool {
return pairs[i][1] < pairs[j][1]
})
ans, cur := 0, math.MinInt32
for _, p := range pairs {
if cur < p[0] {
cur = p[1]
ans++
}
}
return ans
}

• function findLongestChain(pairs: number[][]): number {
pairs.sort((a, b) => a[1] - b[1]);
let res = 0;
let pre = -Infinity;
for (const [a, b] of pairs) {
if (pre < a) {
pre = b;
res++;
}
}
return res;
}


• impl Solution {
pub fn find_longest_chain(mut pairs: Vec<Vec<i32>>) -> i32 {
pairs.sort_by(|a, b| a[1].cmp(&b[1]));
let mut res = 0;
let mut pre = i32::MIN;
for pair in pairs.iter() {
let a = pair[0];
let b = pair[1];
if pre < a {
pre = b;
res += 1;
}
}
res
}
}